Question

if 270°<A<360°,90°<B<180°,cos A=5/13,tan B=-15/8,then sin(A+B)


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Answer 1

Answer:

cosA=base/hypotuse=5/13

perpendicular=169-25=144

$\sqrt{144 } = 12$

perpendicular is 12

tanB=15/8 =perpendicular/base

so hypotuse for B is 17

So sin ( A+B)=sinAcosB+ cosAsinB

sinA= 12/13

sinB=15/17

cosA=5/13

cosB=8/17

=12/13*8/17+5/13*15/17

=96/221+75/221

=171/221

Answer 2

Hi there

Here is your answer

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270°<A<360        =  4 quadrant                                                  

90°<B<180°          = 2 quadrant

Given,   cos A=5/13       tan B=-15/8

In 4 quadrant  sin is negative

cos A=base/hypotenuse = 5/13

sin A= perpendicular/hypotenuse =12/13    {By using Pythagoras theoram}

but in 4 quadrant sin is negative= -12/13

Similarly,

tan B=perpendicular/base=-15/8

sin B= perpendicular/hypotenuse =-15/17

but in 2 quadrant sin is positive= 15/17

similarly cos B=-8/17  {cos is negative in 2 quadrant}

We know that  sin(A+B)=sinAcosB +cosAsinB

By putting values we get,

sin(A+B)= -12/13 × -8/17 + 5/13 × 15/17

             = 96/221 + 75/221

          Ans =171/221

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